Representing and Manipulating Floating Points Jin-Soo Kim (jinsookim@skku.edu) Computer Systems Laboratory Sungkyunkwan University http://csl.skku.edu
The Problem How to represent fractional values with finite number of bits?.1.612 3.1415926535897932384626433832795288... Wide ranges of numbers 1 Light-Year = 9,46,73,472,58.8 km The radius of a hydrogen atom:.25 m 2
Fractional Binary Numbers (1) Representation Bits to right of binary point represent fractional powers of 2 i Represents rational number: b k 2 k 2 i 2 i 1 k j 4 2 1 b i b i 1 b 2 b 1 b. b 1 b 2 b 3 b j 1/2 1/4 1/8 2 j 3
Fractional Binary Numbers (2) Examples: Value Representation 5-3/4 11.11 2 2-7/8 1.111 2 63/64.111111 2 Observations Divide by 2 by shifting right Multiply by 2 by shifting left Numbers of form.111111.. 2 just below 1. 1/2 + 1/4 + 1/8 + + 1/2 i + 1. Use notation 1. 4
Fractional Binary Numbers (3) Representable numbers Can only exactly represent numbers of the form x / 2 k Other numbers have repeating bit representations Value Representation 1/3.11111[1] 2 1/5.111111[11] 2 1/1.111111[11] 2 5
Fixed-Point Representation (1) p.q Fixed-point representation Use the rightmost q bits of an integer as representing a fraction Example: 17.14 fixed-point representation 1 bit for sign bit 17 bits for the integer part 14 bits for the fractional part An integer x represents the real number x / 2 14 Maximum value: (2 31 1) / 2 14 13171.999 Integer part (2 2 +2 1 =5) Fractional part (2-1 +2-2 +2-4 =.8125) 1 1 1 1 1 Sign bit Imaginary binary point 6
Fixed-Point Representation (2) Properties Convert n to fixed point: n * f (= n << q) Add x and y: x + y + 1 1 1 1 1 1 1 13.5 1.75 1 1 1 1 1 15.25 Subtract y from x: Add x and n: Multiply x by n: Divide x by n: x y x + n * f x * n x / n x, y: fixed-point number n: integer f = 1 << q 7
Fixed-Point Representation (3) Pros Simple Can use integer arithmetic to manipulate No floating-point hardware needed Used in many low-cost embedded processors or DSPs (digital signal processors) Cons Cannot represent wide ranges of numbers 8
Representing Floating Points IEEE standard 754 Established in 1985 as uniform standard for floating-point arithmetic Before that, many idiosyncratic formats Supported by all major CPUs William Kahan, a primary architect of IEEE 754, won the Turing Award in 1989 Driven by numerical concerns Nice standards for rounding, overflow, underflow Hard to make go fast Numerical analysts predominated over hardware types in defining standard 9
FP Representation Numerical form: -1 s x M x 2 E Sign bit s determines whether number is negative or positive Significand M normally a fractional value in range [1., 2.) Exponent E weights value by power of two Encoding s exp frac MSB is sign bit s exp field encodes E (Exponent) frac field encodes M (Mantissa) 1
FP Precisions s exp frac Single precision 8 exp bits, 23 frac bits (32 bits total) Double precision 11 exp bits, 52 frac bits (64 bits total) Extended precision 15 exp bits, 63 frac bits Only found in Intel-compatible machines Stored in 8 bits (1 bit wasted) 11
Normalized Values Condition: exp and exp 111 1 Exponent coded as a biased value E = Exp Bias Exp: unsigned value denoted by exp Bias: Bias value (=2 k-1-1, where k is the number of exp bits) Single precision (k=8): 127 (Exp: 1..254, E: -126..127) Double precision (k=11): 123 (Exp: 1..246, E: -122..123) Significand coded with implied leading 1 M = 1.xxx x 2 Minimum when frac = (M = 1.) Maximum when frac = 111 1 (M = 2. ) Get extra leading bit for free 12
Normalized Values: Example float f = 23.; 23 1 = 11111111 2 = 1.1111111 2 X 2 1 Significand M = 1.1111111 2 frac = 1111111 2 Exponent E = 1 Exp = E + Bias = 1 + 127 = 137 = 111 2 Hex: 4 4 F A 6 Binary: 1 1 1111 11 11 137: 1 1 1 23: 1111 11 11 13
DenormalizedValues Condition: exp = Value Exponent value E = 1 Bias Significand value M =.xxx x 2 (no implied leading 1) Case 1: exp =, frac = Represents value. Note that there are distinct values + and - Case 2: exp =, frac Numbers very close to. Gradual underflow : possible numeric values are spaced evenly near. 14
Condition: exp = 111 1 Special Values Case 1: exp = 111 1, frac = Represents value (infinity) Operation that overflows Both positive and negative e.g. 1./. = 1./. = +, 1./. = Case 2: exp = 111 1, frac Not-a-Number (NaN) Represents case when no numeric value can be determined e.g. sqrt( 1),, *, 15
Tiny FP Example (1) 8-bit floating point representation The sign bit is in the most significant bit The next four bits are the exp, with a bias of 7 The last three bits are the frac Same general form as IEEE format Normalized, denormalized Representation of, NaN, infinity 7 6 3 2 s exp frac 16
Tiny FP Example (2) Values related to the exponent (Bias = 7) Description Exp exp E = Exp - Bias 2 E Denormalized -6 1/64 Normalized 1 2 3 4 5 6 7 8 9 1 11 12 13 14 1 1 11 1 11 11 111 1 11 11 111 11 111 111 inf, NaN 15 1111 - - -6-5 -4-3 -2-1 1 2 3 4 5 6 7 1/64 1/32 1/16 1/8 1/4 1/2 1 2 4 8 16 32 64 128 17
Dynamic range Tiny FP Example (3) Description Bit representation e E f M V Zero -6 Smallest pos. 1-6 1/8 1/8 1/512 1-6 2/8 2/8 2/512 11-6 3/8 3/8 3/512 11-6 6/8 6/8 6/512 Largest denorm. 111-6 7/8 7/8 7/512 Smallest norm. 1 1-6 8/8 8/512 1 1 1-6 1/8 9/8 9/512 11 11 11 111 6 6-1 -1 6/8 7/8 14/8 15/8 14/16 15/16 One 111 7 8/8 1 111 1 7 1/8 9/8 9/8 111 1 7 2/8 1/8 1/8 Largest norm. 111 11 111 111 14 14 7 7 6/8 7/8 14/8 15/8 224 24 Infinity 1111 - - - - + 18
Tiny FP Example (4) Encoded values (nonnegative numbers only) 111 XXX = (8/8 ~ 15/8)*2 6 111 XXX = (8/8 ~ 15/8)*2 7 16 32 48 64 8 96 112 128 144 16 176 192 28 224 24 256 + 111 XXX = (8/8 ~ 15/8)*2 1 XXX = (8/8 ~ 15/8)*2 1..5 1. 1.5 2. 2.5 3. 3.5 4. XXX = (/8 ~ 7/8)*2-6 1 XXX = (8/8 ~ 15/8)*2-6 11 XXX = (8/8 ~ 15/8)*2-4..2.4.6.8.1.12 (Without denormalization)..2.4.6.8.1.12 XXX = (8/8 ~ 15/8)*2-7 19
Interesting Numbers Normalized Denorm +Denorm +Normalized + NaN + NaN Description exp frac Numeric Value Zero. Smallest Positive denormalized Largest Denormalized Smallest Positive Normalized 1 111 11 1 Single: 2-23 X 2-126 1.4 X 1-45 Double: 2-52 X 2-122 4.9 X 1-324 Single: (1. ε) X 2-126 1.18 X 1-38 Double: (1. ε) X 2-122 2.2 X 1-38 Single: 1. X 2-126, Double: 1. X 2-122 (Just larger than largest denormalized) One 11 11 1. Largest Normalized 111 1 111 11 Single: (2. ε) X 2 127 3.4 X 1 38 Double: (2. ε) X 2 123 1.8 X 1 38 2
Special Properties FP zero same as integer zero All bits = Can (almost) use unsigned integer comparison Must first compare sign bits Must consider = NaNs problematic Will be greater than any other values Otherwise OK Denormalized vs. normalized Normalized vs. Infinity 21
Rounding For a given value x, finding the closest matching value x that can be represented in the FP format IEEE 754 defines four rounding modes Round-to-even avoids statistical bias by rounding upward or downward so that the least significant digit is even Rounding modes $1.4 $1.6 $1.5 $2.5 $ 1.5 Round-toward-zero $1 $1 $1 $2 $ 1 Round-down ( ) $1 $1 $1 $2 $ 2 Round-up (+ ) $2 $2 $2 $3 $ 1 Round-to-even (default) or Round-to-nearest $1 $2 $2 $2 $ 2 22
Round-to-Even Round up conditions R = 1, S = 1 >.5 G = 1, R = 1, S = Round to even 1.BBGRXXX Guard bit: LSB of result Round bit: 1 st bit removed Sticky bit: OR of remaining bits Value Fraction GRS Up? Rounded 128 1. (x2 7 ) No 1. 13 1.11 (x2 3 ) 1 No 1.11 17 1.1 (x2 4 ) 1 No 1. 19 1.11 (x2 4 ) 11 Yes 1.1 138 1.11 (x2 7 ) 11 Yes 1.1 63 1.11111 (x2 5 ) 111 Yes 1. 23
FP Addition Adding two numbers: (Assume E1 > E2) Align binary points Shift right M2 by E1 E2 Add significands + Result: Sign s, Significand M, Exponent E (= E1) Normalize result if (M 2), shift M right, increment E s1 E1 M1 + s2 E2 M2 ( 1) s1 M1 if (M < 1), shift M left k positions, decrement E by k Check for overflow (E out of range?) Round M and renormalize if necessary ( 1) s M E1 E2 ( 1) s2 M2 24
FP Multiplication Multiplying two numbers: Obtain exact result Sign s = s1 ^ s2 Significand M = M1 x M2 Exponent E = E1 + E2 s1 E1 M1 x s2 E2 M2 The biggest chore is multiplying significands Normalize result if (M 2), shift M right, increment E if (M < 1), shift M left k positions, decrement E by k Check for overflow (E out of range?) Round M and renormalize if necessary 25
Floating Points in C C guarantees two levels float (single precision) vs. double (double precision) Conversions double or float int Truncates fractional part Like rounding toward zero Not defined when out of range or NaN (Generally sets to TMin) int double Exact conversion, as long as int has 53 bit word size int float Will round according to rounding mode 26
FP Example 1 #include <stdio.h> int main () { int n = 123456789; int nf, ng; float f; double g; } f = (float) n; g = (double) n; nf = (int) f; ng = (int) g; printf ( nf=%d ng=%d\n, nf, ng); 27
FP Example 2 #include <stdio.h> int main () { double d; d = 1. +.1 +.1 +.1 +.1 +.1 +.1 +.1 +.1 +.1 +.1; } printf ( d = %.2f\n, d); 28
FP Example 3 #include <stdio.h> int main () { float f1 = (3.14 + 1e2) 1e2; float f2 = 3.14 + (1e2 1e2); } printf ( f1 = %f, f2 = %f\n, f1, f2); 29
Ariane 5 Ariane 5 tragedy (June 4, 1996) Exploded 37 seconds after liftoff Satellites worth $5 million Why? Computed horizontal velocity as floatingpoint number Converted to 16-bit integer Careful analysis of Ariane 4 trajectory proved 16-bit is enough Reused a module from 1-year-old software Overflowed for Ariane 5 No precise specification for the software 3
Summary IEEE floating point has clear mathematical properties Represents numbers of form M x 2 E Can reason about operations independent of implementation As if computed with perfect precision and then rounded Not the same as real arithmetic Violates associativity / distributivity Makes life difficult for compilers and serious numerical applications programmers 31